Computer Science Canada

odd indices of an array
Author:  Krocker [ Sun Oct 20, 2013 5:13 pm ]
Post subject:  odd indices of an array
hi, so i need to output the odd number indices of an array without using any if statments. i know that odd = 2n+1 where n is the an integer starting at 0. my trouble is getting the for loop im using to stop when all the odd indices are outputed. So far, all i get it to is before the last index or beyond the last index, giving me an out of bound error. This is my for loop so far:

code:

System.out.println ("\n\nValues at odd-numbered indices:");
                int odd = 2;
                int shiftedOddArray = 0;
               
                for (int i = 0; i < (odd-1); i++){
                        odd = (2*i)+1;
                        shiftedOddArray = shiftedArray[odd];
                        System.out.print ( shiftedOddArray + "\t");
                }


shiftedArray is the original array that i have to output every odd index. The number of indices and the numbers in the indices are all determined by the user at the begining of the program.
Author:  Raknarg [ Sun Oct 20, 2013 7:47 pm ]
Post subject:  RE:odd indices of an array
You don't have to use i++. THat's just a shortcut for saying add one to i. You can do whatever you want in there (I'm pretty sure)
Author:  Tony [ Sun Oct 20, 2013 8:48 pm ]
Post subject:  RE:odd indices of an array
anything you want, or even nothing.
code:

for(;;){
   ...
}

is a valid loop. At least for C, probably Java.
Author:  Krocker [ Sun Oct 20, 2013 9:42 pm ]
Post subject:  RE:odd indices of an array
but how would that stop when the program has gone through the number of odd indices that are possiible according tho the user specified number of indices?
Author:  Insectoid [ Sun Oct 20, 2013 10:13 pm ]
Post subject:  RE:odd indices of an array
How would you do it if you wanted to run through all of the indices of the array up to the user specified limit?
Author:  Raknarg [ Mon Oct 21, 2013 8:05 am ]
Post subject:  RE:odd indices of an array
@Tony is that a loop that does nothing or loops infinitely?
Author:  Insectoid [ Mon Oct 21, 2013 8:14 am ]
Post subject:  RE:odd indices of an array
The 2nd argument to a for loop is the exit condition. If there is no exit condition, what do you think will happen?
Author:  Krocker [ Mon Oct 21, 2013 8:36 am ]
Post subject:  RE:odd indices of an array
if there isnt an exit condition, there will be an infinite loop, thats not what im looking for. Anyways, i figured it out, i simply used a while loop and said that while (odd< numberOfArray)
Author:  C14 [ Tue Oct 22, 2013 1:04 pm ]
Post subject:  Re: RE:odd indices of an array
Krocker @ Mon Oct 21, 2013 8:36 am wrote:
Anyways, i figured it out, i simply used a while loop and said that while (odd< numberOfArray)


Can we see your code?
Author:  Krocker [ Fri Oct 25, 2013 8:28 am ]
Post subject:  Re: odd indices of an array
Heres the code that i used:

code:

                System.out.println ("\n\nValues at odd-numbered indices:");
                int odd = 1; //declare and initializes variables to be used
                int shiftedOddArray = 0;
                int n = 0;

                //does through each odd indices and outputs the value in order of the shifted array by calling shiftedArray [odd] where odd is 2n+1
                while (odd <= numberOfArray-1){
                        shiftedOddArray = shiftedArray[odd];
                        System.out.print ( shiftedOddArray + "\t");
                        n++;
                        odd = (2*n)+1;
                }

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